Roots of Unity: Finding Primitive Root W6 and Solving for 1/W6

[fredrogers3]

Homework Statement

Hello everyone,

In this problem, I was to mark all the sixth roots of 1 in the complex plane. Then, I was to figure out what the primitive root W₆ is.

However, I am stuck by the question: "Which power of W₆ is equal to 1/W₆?"

Homework Equations

See Below

The Attempt at a Solution

In this case, W₆ appears to be 1, so could I just not use the power n=-6 to do this?
This would yield me the value cos(-2*pi) which would be equal to 1. Thus, in general, for W_n, do I simply multiply by -n to get the reciprocal?

Thanks

 

[Mark44]

Homework Statement

Hello everyone,

In this problem, I was to mark all the sixth roots of 1 in the complex plane. Then, I was to figure out what the primitive root W₆ is.

However, I am stuck by the question: "Which power of W₆ is equal to 1/W₆?"

Homework Equations

See Below

The Attempt at a Solution

In this case, W₆ appears to be 1, so could I just not use the power n=-6 to do this?

What does a power of -6 have to do with this problem? The question is asking whichy root is equal to its own reciprocal.

When you say "W₆ appears to be 1" did you do as the problem asks, and mark each of the six roots?

This would yield me the value cos(-2*pi) which would be equal to 1.

Yes, but cos(2$\pi$) is also equal to 1. Why do you have the number -2 multiplying pi?

Thus, in general, for W_n, do I simply multiply by -n to get the reciprocal?

?
From your question, I'm guessing that you did NOT mark the six sixth roots of 1. If you have all six of them drawn on the unit circle, it should be obvious that two of them are their own reciprocals.

 

[fredrogers3]

What does a power of -6 have to do with this problem? The question is asking whichy root is equal to its own reciprocal.

When you say "W₆ appears to be 1" did you do as the problem asks, and mark each of the six roots?
Yes, but cos(2$\pi$) is also equal to 1. Why do you have the number -2 multiplying pi?

?
From your question, I'm guessing that you did NOT mark the six sixth roots of 1. If you have all six of them drawn on the unit circle, it should be obvious that two of them are their own reciprocals.

I have W₃=-1 and W₆=1, so these two equal their own reciprocal. Is that what the question is asking? I seem to have completely misunderstood the question.

 

[Mark44]

That's what I think the problem is asking for, but the problem statement isn't as clear as it should be, IMO. A clearer problem statement would be this:
Mark the six sixth roots of 1 as points on the unit circle in the complex plane. Which of these roots is its own reciprocal?
Answer to the second question:
w₀ = 1, (w₀)⁶ = 1, so w₀ is a sixth root of 1, and 1/w₀ = 1

w₃ = -1, (w₃)⁶ = 1, so w₃ is another sixth root of 1, and 1/w₃ = -1

I'm not sure how they are numbering the roots. I am numbering them w₀ (= 1), w₁, ..., w₅. If they are numbering them w₁, w₂,..., w₆, it's hard to say what w₆ is if I don't know what they're calling w₁.

 

[haruspex]

I interpret the notation rather differently. I believe the author is writing ω_n for the principal n^th root of unity. The 6 roots of unity are therefore, by definition, the powers of ω₆, ω₆ having the least nonzero argument of them. ω_n for n≠6 is irrelevant to the question.